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Difference between revisions of "2017 AMC 10A Problems/Problem 14"

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If we replace <math>A</math> with <math>100</math> we get a system of equations, and the sum of the values of <math>M</math> and <math>S</math> is the percentage of <math>A</math>.
 
If we replace <math>A</math> with <math>100</math> we get a system of equations, and the sum of the values of <math>M</math> and <math>S</math> is the percentage of <math>A</math>.
 
Solving, we get <math>S=\frac{400}{99}</math> and <math>M=\frac{1900}{99}</math>.
 
Solving, we get <math>S=\frac{400}{99}</math> and <math>M=\frac{1900}{99}</math>.
−
Adding, we get <math>\frac{2300}{99}</math>, which is closest to <math>23</math> which is <math>(D)</math>.
+
Adding, we get <math>\frac{2300}{99}</math>, which is closest to <math>23</math> which is <math>\boxed{\textbf{(D)}\ 23\%}</math>.
  
 
-Harsha12345
 
-Harsha12345
Line 59: Line 59:
 
~mathboy282
 
~mathboy282
  
 +
==Solution 4==
 +
Let <math>m</math> be the price of a movie ticket and <math>s</math> be the price of a soda.
 +
 +
Then,
 +
 +
<cmath>m=\frac{A-s}{5}</cmath>
 +
and
 +
<cmath>s=\frac{A-m}{20}</cmath>
 +
Then, we can turn this into
 +
<cmath>5m=A-s</cmath>
 +
<cmath>20s=A-m</cmath>
 +
 +
Subtracting and getting rid of A, we have <math>20s-5m=-m+s \rightarrow 19s=4m</math>. Assume WLOG that <math>s=4</math>, <math>m=19</math>, thus making a solution for this equation. Substituting this into the 1st equation, we get <math>A=99</math>. Hence, <math>\frac{m+s}{A} = \frac{19+4}{99} \approx \boxed{\textbf{(D)}\ 23\%}</math>
 +
 +
~MrThinker
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/s4vnGlwwHHw
 
https://youtu.be/s4vnGlwwHHw

Revision as of 21:47, 25 October 2022

Problem

Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was $A$ dollars. The cost of his movie ticket was $20\%$ of the difference between $A$ and the cost of his soda, while the cost of his soda was $5\%$ of the difference between $A$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $A$ did Roger pay for his movie ticket and soda?

$\mathrm{\textbf{(A)} \ }9\%\qquad \mathrm{\textbf{(B)} \ } 19\%\qquad \mathrm{\textbf{(C)} \ } 22\%\qquad \mathrm{\textbf{(D)} \ } 23\%\qquad \mathrm{\textbf{(E)} \ }25\%$

Solution

Let $m$ = cost of movie ticket
Let $s$ = cost of soda

We can create two equations:

\[m = \frac{1}{5}(A - s)\] \[s = \frac{1}{20}(A - m)\]

Substituting we get:

\[m = \frac{1}{5}(A - \frac{1}{20}(A - m))\] which yields:
\[m = \frac{19}{99}A\]

Now we can find s and we get:

\[s = \frac{4}{99}A\]

Since we want to find what fraction of $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:

\[\frac{19}{99}A + \frac{4}{99}A \implies \boxed{\textbf{(D)}\ 23\%}\]

Solution 2

We have two equations from the problem: $5M=A-S$ and $20S=A-M$ If we replace $A$ with $100$ we get a system of equations, and the sum of the values of $M$ and $S$ is the percentage of $A$. Solving, we get $S=\frac{400}{99}$ and $M=\frac{1900}{99}$. Adding, we get $\frac{2300}{99}$, which is closest to $23$ which is $\boxed{\textbf{(D)}\ 23\%}$.

-Harsha12345

Solution 3

WLOG let $A=20.$ Let $m$ be the price of the movie ticket. Let $s$ be the price of the soda. Thus, \begin{align*} m &=\frac{1}{5}\left(20-s\right) \\ s &= \frac{1}{20}\left(20-m\right) \end{align*} Simplifying, we have \begin{align*} 5m &= 20 - s \\ 20s &= 20-m \end{align*}

Multiplying the first equation by $4$ and adding them, we have \[m+s = \frac{100 - 4s - m}{20}\]

Finding $m$ and $s$ is straightforward from there.

~mathboy282

Solution 4

Let $m$ be the price of a movie ticket and $s$ be the price of a soda.

Then,

\[m=\frac{A-s}{5}\] and \[s=\frac{A-m}{20}\] Then, we can turn this into \[5m=A-s\] \[20s=A-m\]

Subtracting and getting rid of A, we have $20s-5m=-m+s \rightarrow 19s=4m$. Assume WLOG that $s=4$, $m=19$, thus making a solution for this equation. Substituting this into the 1st equation, we get $A=99$. Hence, $\frac{m+s}{A} = \frac{19+4}{99} \approx \boxed{\textbf{(D)}\ 23\%}$

~MrThinker

Video Solution

https://youtu.be/s4vnGlwwHHw

https://youtu.be/zY726PV6XU8

~savannahsolver

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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